= Solution
The second <moment> is $E[X^2]=\alpha\int_1^\infty x^{1-\alpha}\,dx$, finite exactly for $\alpha>2$, where it equals $\alpha/(\alpha-2)$. Subtracting the square of the <mean> gives
$$
\boxed{\operatorname{Var}(X)=\frac{\alpha}{\alpha-2}
-\frac{\alpha^2}{(\alpha-1)^2}
=\frac{\alpha}{(\alpha-1)^2(\alpha-2)},\qquad\alpha>2}.
$$
For $1<\alpha\leq2$, the <mean> exists but the <variance> is infinite. For $0<\alpha\leq1$, the ordinary centered <variance> is not defined because the <mean> is not finite. Thus there is no finite <variance> for any $\alpha\leq2$.
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